Question: What is the value of $\dfrac{d}{dx}\sec(x)$ at $x=\dfrac{11\pi}{6}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac23$ (Choice B) B ${\sqrt{3}}$ (Choice C) C $-\dfrac{1}{\sqrt{3}}$ (Choice D) D $\dfrac32$
Let's first find $\dfrac{d}{dx}\sec(x)$. Then, we can evaluate it at $x=\dfrac{11\pi}{6}$. Recall that the derivative of $\sec(x)$ is $\dfrac{\sin(x)}{\cos^2(x)}$, or $\sec(x)\tan(x)$. Put another way, $\dfrac{d}{dx}[\sec(x)]=\dfrac{\sin(x)}{\cos^2(x)}=\sec(x)\tan(x)$. [Is there a way to know this without memorizing?] Now let's plug in $x={\dfrac{11\pi}{6}}$ : $\begin{aligned} &\phantom{=}\dfrac{\sin\left({\dfrac{11\pi}{6}}\right)}{\cos^2\left({\dfrac{11\pi}{6}}\right)} \\\\ &=\dfrac{-\dfrac12}{\left(\dfrac{\sqrt{3}}{2}\right)^2} \\\\ &={-\dfrac12}\cdot \dfrac{4}{3} \\\\ &=-\dfrac23 \end{aligned}$ In conclusion, the value of $\dfrac{d}{dx}\sec(x)$ at $x=\dfrac{11\pi}{6}$ is $-\dfrac23$.